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As mentioned in the text, the notation∃!xP(x) denotes "There exists a unique x such that P(x)is true." If the domain consists of all integers, what are the truth values of these statements? a) ∃!x(x > 1) b) ∃!x(x2 =1) c) ∃!x(x+3=2x) d) ∃!x(x = x +1)

Respuesta :

Answer:

The third statement is true here.

Step-by-step explanation:

a)

If we take some random values like 5 and 10, it is clear that both 5 and 10 is greater than 1. Hence, there is no such unique x, for which P(x) > 1 is true. We can find multiple values which will be greater than 1.

b)

For both the values of x, 1 and -1, the value of [tex]x^{2}[/tex] is 1. In this case, we are getting at least two values of x, for which P(x) is true.

Hence, this statement is not true.

c)

[tex]x + 3 = 2x\\x = 3[/tex].

In this scenario, we are getting an unique value of x that is 3, for which P(x) is true.

This statement is true.

d)

[tex]x = x + 1\\0 = 1[/tex].

The above one is not an equation.

There is no meaning of the above.

Hence, this statement can not be true.

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